已知cos(α-π/4)=根号2/10,α∈(π/2,π)

1.cosα-sinα=√2·[sin(π/4)·cosα-cos(π/4)·sinα]=√2·sin(π/4 -α)=-√2·sin(α -π/4)
由于α∈(π/2,π),故α -π/4∈(π/4,3π/4);则sin(α -π/4)>0.由cos(α-π/4)=根号2/10得到sin(α -π/4)=7√2/10.
则cosα-sinα=-7/5.
2.
cos(2α+π/3）= sin[(2α-π/2)+5π/6] = -sin[(2α-π/2)-π/6] = -2·sin[(α-π/4) - π/12]·cos[(α-π/4) - π/12]
= -2·[sin(α -π/4)·cos(π/12) -cos(α -π/4)·sin(π/12)]·[cos(α -π/4)·cos(π/12) +sin(α -π/4)·sin(π/12)]
= -2·[(7√2/10)·cos(π/12) -(√2/10)·sin(π/12)]·[(√2/10)·cos(π/12) +(7√2/10)·sin(π/12)]
= -2·[(7/5)·cos^2(π/12) +(48/5)·sin(π/12)·cos(π/12) -(7/5)·sin^2(π/12)]
= -2·[(7/5)·(cos^2(π/12) - sin^2(π/12) ) +(24/5)·sin(π/6)]
= -2·[(7/5)·cos(π/6) +(24/5)·sin(π/6)]
= -2·[(7/5)·(√3/2) +(24/5)·(1/2)]
= -(7√3 +24)/5